# Exponential and Logarithmic problems

• Jul 24th 2007, 10:23 AM
JonathanEyoon
Exponential and Logarithmic problems
I can do such problems like

6^x+1 = 4^2x-1

but I can't figure out how to do

.05(1.15)^X = 5

Help =(
• Jul 24th 2007, 10:29 AM
galactus
$\frac{1}{20}(\frac{23}{20})^{x}=5$

$(\frac{23}{20})^{x}=100$

$xln(\frac{23}{20})=ln(100)$

$x=\frac{2ln(10)}{ln(\frac{23}{20})}$

$\approx{32.95}$
• Jul 24th 2007, 10:32 AM
JonathanEyoon
Quote:

Originally Posted by galactus
$\frac{1}{20}(\frac{23}{20})^{x}=5$

$(\frac{23}{20})^{x}=100$

$xln(\frac{23}{20})=ln(100)$

$x=\frac{2ln(10)}{ln(\frac{23}{20})}$

$\approx{32.95}$

Where did the 100 on the other side come from?
• Jul 24th 2007, 12:15 PM
Krizalid
It's really simple


\begin{aligned}
\frac{1}
{{20}}\left( {\frac{{23}}
{{20}}} \right)^x &= 5\\
20 \cdot \frac{1}
{{20}}\left( {\frac{{23}}
{{20}}} \right)^x &= 20 \cdot 5\\
\left( {\frac{{23}}
{{20}}} \right)^x &= 100
\end{aligned}
• Jul 24th 2007, 12:20 PM
Krizalid
Quote:

Originally Posted by JonathanEyoon
6^x+1 = 4^2x-1

I assume you forgot the brackets


\begin{aligned}
6^{x + 1} &= 4^{2x - 1}\\
(x + 1)\log 6 &= 2(2x - 1)\log 2\\
x\log 6 + \log 6 &= 4x\log 2 - 2\log 2\\
4x\log 2 - x\log 6 &= 2\log 2 + \log 6\\
x(4\log 2 - \log 6) &= 2\log 2 + \log 6\\
x &= \frac{{2\log 2 + \log 6}}
{{4\log 2 - \log 6}}
\end{aligned}